TITLE: What is the Proof that this Diagram is not a Category?
QUESTION [6 upvotes]: Reading through the Category Theory Wikibook, I came across the following exercise:

(Harder.) If we add another morphism to the above example, it fails to
  be a category. Why? Hint: think about associativity of the composition
  operation.
  

I haven't been able to figure out why the diagram given is not a category. The circles are the objects, the arrows are the morphisms, and there is a clear notion of composition.
Going through the category laws (in reverse order):

1. There is an identity element for every object ($id_A$ and
  $id_B$).
2. The set of morphisms are closed under composition: $f \circ g
> = id_A$, $h \circ g = id_A$, $g \circ f = id_B$, $g \circ h = id_B$, and any composition involving an identity morphism is clearly (I
  think) closed. Compositions like $f \circ h$ are meaningless.  
3. Coming to the hint in the prompt, and trying to prove associativity, $h \circ (g \circ f) = g \circ id_B = g$, whereas $(h
> \circ g) \circ f = id_B \circ f = f$, so $h \circ (g \circ f) \neq (h
> \circ g) \circ f$.

This is where I get a little confused, though, since $f$ and $h$ are in some ways "the same", since A and B are isomorphic (right?). I haven't figured out exactly why I'm confused about this, but it seems like this exercise means that there can only ever be a single morphism between two objects in a category, which is clearly nonsense. What am I missing in all this?
Thanks!

REPLY [6 votes]: Well the statement of the problem is not extremely clear, perhaps on purpose, but what it is trying to teach is the following: in the "example above" you conclude - by looking at the diagram without the $h$- that the composition of $f$ and $g$ gives an identity (on $A$ and on $B$). If they now tell you to add another - different - morphism $h$, you must be sure that the compositions $g \circ h$ and $h \circ g$ exists. Since the only appropriate and available morphisms are the identities, you are obliged to write: $g \circ h=id_B$, but then 
$$g \circ h=id_B=g \circ f$$ and , by left-composing with $f$ you get the contradiction $f=h$. Indeed $$f\circ g \circ h=id_A \circ h=h=f \circ id_B=f=f \circ g \circ f$$ So you cannot add a different morphism $h$ to the category in the "example above". Put it in another way: you cannot have two different 2-sided inverses ($f$ and $h$) to a morphism $g$.